4 Answers
4

The sum can be written as
$$\sum_{n=0}^\infty \binom{4n}{n}\frac{1}{(3n+1)2^{4n+1}}=\frac{1}{2}{\;}_3F_2 \left(\frac{1}{4},\frac{1}{2},\frac{3}{4}; \frac{2}{3},\frac{4}{3};\frac{16}{27} \right)$$
For the Hypergeometric Function, see equation (25) of http://mathworld.wolfram.com/HypergeometricFunction.html
$${\;}_3F_2 \left(\frac{1}{4},\frac{1}{2},\frac{3}{4}; \frac{2}{3},\frac{4}{3};y \right) = \frac{1}{1-x^3}$$
where $x$ and $y$ are related by
$$y=\left( \frac{4x(1-x^3)}{3}\sqrt[3]{4} \right)^3$$
Putting $y=16/27$, we get a polynomial equation in $x$.
$$1=16 x^3(1-x^3)^3$$
According to Mathematica, this equation has only two real roots which are
$$\begin{align*}
\alpha &= \frac{1}{\sqrt[3]{2}} \\
\beta &= \left\{\frac{1}{6} \left(5-\frac{4}{\left(19-3 \sqrt{33}\right)^{1/3}}-\left(19-3 \sqrt{33}\right)^{1/3}\right) \right\}^{1/3} \end{align*}$$
Numericaly, $\beta$ gives the right answer. So, the final result is
$$ \begin{align*}\sum_{n=0}^\infty \binom{4n}{n}\frac{1}{(3n+1)2^{4n+1}} &=\frac{1}{2\left(1-\beta^3 \right)} \\
&= \dfrac{\sqrt[3]{17+3\sqrt{33}}}{3}-\dfrac{2}{3\sqrt[3]{17+3\sqrt{33}}}-\dfrac{1}{3} \\ &\approx 0.543689\end{align*}$$

The last sum is the Lagrange-Burmann inversion formula applied to $\phi(z)=(1+z)^4$. It is straightforward now that the series (exlcuding the constant factor $1/2^{1/3}$) is one of the two real roots of the quartic $z^4 + z + 16^{-1/3}$. One can exclude $-1/2^{1/3}$ via invoking a numerical check, and thus